4

Let $\beta$ be the surface area of the unit sphere. The Poisson Kernel for the upper half-space $\mathbb{R}_n^+=\left\{x\in\mathbb{R}^n \mid x_n > 0\right\}$ is $$ K\left(x,y\right)=\frac{2x_n}{\beta}\frac{1}{\left|x-y\right|^n}. $$ Its integral w.r.t. $y$ on the boundary of the half-space, $\int_{\partial\mathbb{R}_n^+} K\left(x,y\right)dy$ should be $1$ for all $x\in\mathbb{R}_+^n$. I am not seeing this. Hints?

user5525
  • 103
  • 1
    Already tried the case $n=2$? Can you integrate $\frac{1}{(x-y)^2}$? – Avitus Jun 13 '13 at 08:42
  • Ah, woops. It seems in my calculations I had forgotten the superscript $n$ and hence was getting a divergent integral. For $n=2$, the integral (ignoring the constant terms) is indeed $\int_{-\infty}^{\infty} \frac{1}{\left(x_1-y_1\right)^2+x_2^2}dy_1 = \pi/x_2$, as desired. – user5525 Jun 13 '13 at 09:16
  • Having said that though, the case of $n$ dimensions does not seem trivial. Will work on tomorrow. – user5525 Jun 13 '13 at 09:22
  • 1
    This is proved in S. Axler, P. Bourdon, and W. Ramey, Harmonic function theory, Springer-Verlag, NY-Berlin-Heidelberg, 1992, Chap. 7, pp. 126-128. The proof uses the Kelvin transform. – user64494 Jun 13 '13 at 12:25

0 Answers0